Question

the solubility product Ag3PO4 is: Ksp = 2.8 x 10^-18. What is the solubility of Ag3PO4 in water, in moles per liter?

Answer 1

Answer : The solubility of $Ag_3PO_4$ in water is, $1.8\times 10^{-5}mol/L$

Explanation :

The solubility equilibrium reaction will be:

$Ag_3PO_4\rightleftharpoons 3Ag^++PO_4^{3-}$

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^{+}]^3[PO_4^{3-}]$

$K_{sp}=(3s)^3\times (s)$

$K_{sp}=27s^4$

Given:

$K_{sp}$ = $2.8\times 10^{-18}$

Now put all the given values in the above expression, we get:

$K_{sp}=27s^4$

$2.8\times 10^{-18}=27s^4$

$s=1.8\times 10^{-5}M=1.8\times 10^{-5}mol/L$

Therefore, the solubility of $Ag_3PO_4$ in water is, $1.8\times 10^{-5}mol/L$